Can I write an expression that enters the quaternion space without expressing the variables associated?

Question

First of all, forgive my little knowledge on the subject.

I can enter the complex space by just using an expression only having real numbers, for example:

f(x)= (-1)^x

as f(x)= (-1)^x - Wolfram|Alpha proves. I didn't express anything dependent on the imaginary unit i.

Can I similarly enter the quaternion space, just by using real or complex numbers on the expression? I mean to not use j and k.

Answer 1

A function is well defined only if we specify the Domain and the Codomain. So, if we write simply $f(x)=(-1)^x$ the function is ill-defined. We have to specify if we want a function $f:\mathbb{R} \rightarrow \mathbb{R} $ or $f:\mathbb{R} \rightarrow \mathbb{C} $ or $f:\mathbb{C} \rightarrow \mathbb{C} $, and all these are different functions.

Look well at the page you have linked. When WA gives the ''Properties as a real function'' write that The Domain and the Codomain are the empty set. This means that such a function ( reals to reals) is not defined.

Obviously you can define a function that has $\mathbb{R}$ as Domain and Codomain on the quaternions, but it is a matter of definition.

Answer 2

From set theory we have that $(\mathbb{H}-\mathbb{R})^\mathbb{R}$ is not empty, so let $f\in (\mathbb{H}-\mathbb{R})^\mathbb{R}$. Then $f$ is the function you want. Furthermore, I have not used any variable outside the complex field.

By the way, any function you find there, even if you don't use any variable outside the complex fields may be restated like this.

Let's suppose that for some $x$ in $\mathbb{R}$, $f(x)=a+bi+cj+dk$ and define $f_2(w)=f(w)$ if $w=x$ and $f_2(w)=a+bi+cj+dk$ otherwise. Then $f_2=f$ and even if I didn't use any variable outside the complex field when defining $f$, I did when defining $f_2$.

Without redefining your restriction "not using any variable outside the complex field" in a formal, exact, and non-ambiguous way, this is not a mathematical question.